# Chemistry Calculations (O level)

## Chemistry calculations (O level)

This post is about answering O level chemistry calculations questions which are part of stoichiometry. To fully understand these calculations you should be familiar with:

• how to use the periodic table
• relative molecular mass
• molar mass
• moles
• concentration
• valency

These questions are compatible with Zimsec O level Chemistry and Combined science.

### Question 1

Using the periodic table calculate the molecular masses of the following compounds:

• Na2Cr2O7
• KMnO4
• KClO3
• (NH4)2SO4
• NaHCO3
• CuCO3
• Fe2(SO4)3

### Solution 1

As we know, the relative molecular mass, Mr of a compound is found by adding the relative atomic masses, Ar of the atoms that make up the compound.

• Mr of Na2Cr2O7 = (23 × 2) + (52 × 2) + (16 × 7) = 262
• Mr of KMnO4 = 39 + 55 + (16 × 4) = 158
• Mr of KClO3 = 39 + 35.5 + (16 × 3) = 122.5
• Mr of (NH4)2SO4 = (14 × 2) + (1 × 8) + 32 + (16 × 4) = 132
• Mr of NaHCO3 = 23 + 1 + 12 + (16 × 3) = 84
• Mr of CuCO3 = 64 + 12 + (16 × 3) = 124
• Mr of Fe2(SO4)3 = (56 × 2) + (32 × 3) + (16 × 12) = 400

### Question 2

Write the formulae formulae for the following compounds and then calculate their molecular or formula mass:

• Water
• Carbon monoxide
• Carbon dioxide
• Sulphur dioxide
• Sulphur trioxide
• Carbon tetrachloride
• Magnesium chloride
• Sodium sulphate
• Beryllium hydroxide

### Solution 2

Something to take note of before answering this question: mono- means 1, di- means 2, tri- means 3, tetra- means 4. For example if a compound’s name ends with dioxide it means that the compound contains 2 oxygen atoms.

• Water → H2O
• Carbon monoxide → CO
• Carbon dioxide → CO2
• Sulphur dioxide → SO2
• Sulphur trioxide → SO3
• Carbon tetrachloride → CCl4

Next comes ionic compounds and to write their formulae we need to consider the valencies of their ions or radicals.

• Magnesium chloride → Mg2+ Cl, then the valency of magnesium = number of chlorine atoms required and valency chlorine = number of magnesium atoms required. Therefore the formula = MgCl2.
• Sodium sulphate → Na+ SO42-. So the valency of the sulphate radical becomes the number of sodium ions and the valency of the sodium ion becomes the number of sulphate radicals in the formula. The formula becomes Na2SO4.
by the same token we can find the formulae of the remaining compounds
• Beryllium hydroxide → Be2+ OH = Be(OH)2

### Question 3

Find the mass of:

• 2 moles of nitrogen dioxide
• 0.5 mole of sodium oxide
• 0.01 mole of water
• 4 moles of sodium chloride
• 1.5 mole of calcium carbonate

### Solution 3

• mass of NO2 = number of moles × molar mass
= 2 × [14 + (16 × 2)] = 2 × (14 + 32)
= 2 × 46
= 92 grams
• mass of Na2O = number of moles × molar mass
= 0.5 × [(23 × 2) + 16] = 0.5 × 46 + 16
= 0.5 × 62
= 31 grams
• mass of H2O = number of moles × molar mass
= 0.01 × [(1 × 2) + 16] = 0.01 × 18
= 0.18 grams
• mass of NaCl = number of moles × molar mass
= 4 × (23 + 35.5)
= 4 × 58.5
= 234 grams
• mass of CaCO3 = number of moles × molar mass
= 1.5 × [40 + 12 + (16 × 3)] = 1.5 × (52 + 48)
= 1.5 × 100
= 150 grams

### Question 4

How many moles are there in the following masses?

• 6 g CuCl2
• 30 g FeSO4

### Solution 4

This question can be answered using the formulae:
$latex number\ of\ moles\ = \frac{mass}{molar\ mass}$

• $latex N = \frac{m}{M_r}$
$latex N\ of\ CuCl_2 = \frac{6}{64+(35.5 \times 2)}$
$latex N\ of\ CuCl_2 = \frac{6}{64+71}$
$latex N\ of\ CuCl_2 = \frac{6}{135}$
= 0.0444 moles
• $latex N = \frac{m}{M_r}$
$latex N\ of\ FeSO_4 = \frac{30}{56+32+(16 \times 4)}$
$latex N\ of\ FeSO_4 = \frac{30}{152}$
= 0.2 moles

### Question 5

Calculate the number of moles of the solute in the following volumes of solution:

• 50 cm3 of 0.5 mol/dm3 solution
• 2 dm3 of 2 mol/dm3 solution

### Solution 5

Before we answer this question let us look at the units. Homogeneity of units is the integral part of any correct formula. For example, if the units of concentration are in mol/dm3 then the volume has to be in dm3 for the concentration formula to be correct. Since 1000 cm3 = 1 dm3 we convert cm3 to dm3 by dividing the volume by 1000.

• Number of moles of solute = Concentration × volume of solution
N = CV
$latex N = \frac{50}{1000} \times 0.5$
= 0.025 moles
• N = CV
N = 2 × 2 (no conversion is necessary in this case because the volume is already in dm3)
N = 4 moles

### Question 6

Find the mass of solute in the following volumes:

• 24 cm3 of 0.1 mol/dm3 NaCl
• 250 cm3 of 11,1 g/dm3 of CaCl2

### Solution 6

In the first case we are given the volume and the molar concentration (mol/dm3), so to find the mass we first have to find the number of moles.

• N = CV
$latex N = \frac{24}{1000} \times 0.1$
= 0.0024 moles
then we convert the moles to mass using the formula:
mass = number of moles × molar mass
m = 0.0024 × (23 + 35.5)
= 0.1404 grams

Now we are given the mass concentration (g/dm3) the formula changes to:

• mass = concentration × volume
m = 11.1 × 0.25
= 2.775 g

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